{"id":382,"date":"2026-07-01T00:37:49","date_gmt":"2026-07-01T00:37:49","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=382"},"modified":"2026-08-24T13:04:18","modified_gmt":"2026-08-24T13:04:18","slug":"electric-field","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electric-field\/","title":{"rendered":"What Is an Electric Field?"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nAn electric field is the invisible region of force surrounding any electric charge, where every point is given the force per unit positive charge that a tiny test charge would feel there. Its strength is found from E = F\/q, and for a single point charge from E = kQ\/r<sup>2<\/sup>, measured in newtons per coulomb (N\/C).\n<\/p><\/div>\n\n<p>Pull a wool jumper over your head on a dry winter morning and your hair drifts up to follow it. Reach for a metal door handle and a tiny spark bites your finger. You never touched the charge that did this \u2014 yet something crossed the empty gap and pushed.<\/p>\n\n<p>That invisible reach is the <strong>electric field<\/strong>. Every charged object is wrapped in one, and it is the field \u2014 not the charge itself \u2014 that does the pushing and pulling on everything nearby. Learn to read it and a huge slice of physics, from lightning to the screen you are reading this on, suddenly clicks into place.<\/p>\n\n<h2>What Is an Electric Field?<\/h2>\n\n<p>Think of the space around a charge as quietly &#8220;loaded&#8221;. Bring nothing into it and nothing happens. Bring in another charge and it is shoved, instantly, without anything visibly touching it. The <strong>electric field<\/strong> is our way of describing that loaded space: a map that tells you, at every point, how hard and in which direction a charge would be pushed.<\/p>\n\n<p>More precisely, the electric field at a point is the <strong>force per unit positive charge<\/strong> that a small test charge would experience there. Double the field and you double the force on any charge you place in it. The field exists whether or not you ever put a charge there to feel it.<\/p>\n\n<p>The idea was Michael Faraday&#8217;s. Rather than imagining charges reaching across a void by magic, he pictured invisible lines of force filling the space \u2014 a &#8220;field&#8221; \u2014 and that picture became one of the most powerful ideas in all of physics.<\/p>\n\n<figure style=\"margin:32px auto;max-width:600px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/Michael_Faraday_sitting_crop.jpg\" alt=\"Michael Faraday, who introduced the electric field concept\" loading=\"lazy\" style=\"width:100%;height:auto;border-radius:4px;\" width=\"1095\" height=\"1448\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Michael Faraday introduced the idea of the field \u2014 invisible lines of force filling the space around a charge.<\/figcaption>\n<\/figure>\n\n<h3>Field, force and direction<\/h3>\n\n<p>An electric field is a <strong>vector<\/strong>: it has a size and a direction at every point. By convention, the direction is the way a <em>positive<\/em> test charge would be pushed. So the field points <strong>away<\/strong> from a positive source charge and <strong>towards<\/strong> a negative one.<\/p>\n\n<p>That sign convention matters. A positive charge dropped into a field accelerates along the field direction; a negative charge (an electron, say) feels a force in exactly the opposite direction.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/electric-field-lines-radiating-outward-positive-point.webp\" width=\"396\" height=\"294\" alt=\"Electric field - Field lines radiating outward from a positive point charge, with a test charge at distance r feeling an outward force F equals qE\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:198px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;\">Field lines stream outward from a positive charge. A test charge +q at distance r feels an outward force F = qE.<\/p>\n\n<h2>The Electric Field Formula<\/h2>\n\n<p>There are <strong>two key formulas<\/strong>, and knowing which is which removes most of the confusion. The first is the <em>definition<\/em>; the second is the field of a single point charge.<\/p>\n\n<p>The defining formula links field to force:<\/p>\n\n<div class=\"pf-formula\">E = F \/ q<\/div>\n\n<ul>\n<li><strong>E<\/strong> \u2014 electric field strength, in newtons per coulomb (N\/C), which is identical to volts per metre (V\/m).<\/li>\n<li><strong>F<\/strong> \u2014 the electrostatic force on the test charge, in newtons (N).<\/li>\n<li><strong>q<\/strong> \u2014 the test charge placed in the field, in coulombs (C).<\/li>\n<\/ul>\n\n<p>Rearranged, this gives the formula you will use most often \u2014 the force on <em>any<\/em> charge sitting in a known field:<\/p>\n\n<div class=\"pf-formula\">F = qE<\/div>\n\n<p>The second formula gives the field produced <em>by<\/em> a point charge, a distance r away:<\/p>\n\n<div class=\"pf-formula\">E = kQ \/ r<sup>2<\/sup><\/div>\n\n<ul>\n<li><strong>E<\/strong> \u2014 electric field strength, in newtons per coulomb (N\/C).<\/li>\n<li><strong>k<\/strong> \u2014 Coulomb&#8217;s constant, \u2248 8.99 \u00d7 10<sup>9<\/sup> N\u00b7m<sup>2<\/sup>\/C<sup>2<\/sup> (often rounded to 9.0 \u00d7 10<sup>9<\/sup>).<\/li>\n<li><strong>Q<\/strong> \u2014 the source charge creating the field, in coulombs (C).<\/li>\n<li><strong>r<\/strong> \u2014 the distance from the source charge to the point, in metres (m).<\/li>\n<\/ul>\n\n<p>Coulomb&#8217;s constant is not arbitrary. It is set by the vacuum permittivity \u03b5<sub>0<\/sub> through k = 1\/(4\u03c0\u03b5<sub>0<\/sub>), and \u03b5<sub>0<\/sub> \u2248 8.854 \u00d7 10<sup>\u221212<\/sup> F\/m is one of the measured constants of nature published by <a href=\"https:\/\/physics.nist.gov\/cuu\/Reference\/contents.html\" target=\"_blank\" rel=\"noopener\">NIST<\/a>.<\/p>\n\n<p>Notice what the second formula does <em>not<\/em> contain: the test charge q. The field around a source charge is the same whether you probe it with a huge charge, a tiny one, or none at all.<\/p>\n\n<p>Once you know the source charge and the distance, you can check the field strength in seconds with our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/electric-field\">Electric Field Calculator<\/a> \u2014 it rearranges E = kQ\/r<sup>2<\/sup> to solve for the field, the charge or the distance.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Use this formula<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">When you know\u2026<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">It gives you\u2026<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>E = F\/q<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the force F on a known test charge q<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the field at that point (the definition)<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>E = kQ\/r<sup>2<\/sup><\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the source charge Q and distance r<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the field around a single point charge<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>F = qE<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the field E and a charge q in it<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the force that charge feels<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>E = V\/d<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the voltage V across plates and gap d<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the uniform field between parallel plates<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h3>Why N\/C and V\/m are the same unit<\/h3>\n\n<p>Newtons per coulomb and volts per metre look unrelated, but they are equal. A volt is a joule per coulomb, and a joule is a newton-metre, so V\/m = (N\u00b7m\/C)\/m = N\/C. Use N\/C when you are thinking about force, and V\/m when you are thinking about voltage across a gap \u2014 they describe the same quantity.<\/p>\n\n<h2>How the Electric Field Formula Works<\/h2>\n\n<p>Where does E = kQ\/r<sup>2<\/sup> actually come from? It falls straight out of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/coulombs-law\/\">Coulomb&#8217;s law<\/a>. Coulomb&#8217;s law gives the force between two point charges, Q and q, as F = kQq\/r<sup>2<\/sup>. The field is just that force shared out per unit test charge:<\/p>\n\n<p>E = F\/q = (kQq\/r<sup>2<\/sup>) \u00f7 q = <strong>kQ\/r<sup>2<\/sup><\/strong>.<\/p>\n\n<p>The q cancels. That single cancellation is the whole reason the field is a property of the source alone \u2014 it is what lets us draw one field map and use it for every charge we later place in it.<\/p>\n\n<p>The same result drops out of Gauss&#8217;s law in one step: enclose the charge in an imaginary sphere of radius r, and the field \u2014 equal everywhere on that sphere by symmetry \u2014 works out to kQ\/r<sup>2<\/sup>. A clean version of this derivation using a spherical Gaussian surface is set out in <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/electric\/elesph.html\" target=\"_blank\" rel=\"noopener\">HyperPhysics<\/a>.<\/p>\n\n<h3>The inverse-square fall-off<\/h3>\n\n<p>The r<sup>2<\/sup> in the denominator is the part students underestimate. Because the field depends on 1\/r<sup>2<\/sup>, distance punishes it hard. Move twice as far from a charge and the field does not halve \u2014 it drops to a <strong>quarter<\/strong>. Move three times as far and it falls to a ninth.<\/p>\n\n<p>This is why a charged comb lifts paper from a centimetre away but does nothing from across the room. A small change in distance is a large change in field.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/electric-field-strength-against-distance-inverse-square.webp\" width=\"396\" height=\"280\" alt=\"Graph of electric field strength against distance showing an inverse-square decay; doubling the distance reduces the field to one quarter\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:198px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;\">The field of a point charge weakens as 1\/r<sup>2<\/sup>: doubling the distance cuts it to one quarter.<\/p>\n\n<p>Try it for yourself \u2014 change the charge and the distance in the lab below and watch the field strength respond:<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Electric Field Lab<\/span><\/div><div class=\"pf-sim-slot-body\">\n<style>\n.pf-sim-frame{\nwidth:100%;\nborder:none;\nheight:600px\n}\n@media(max-width:760px){\n.pf-sim-frame{\nheight:1000px\n}\n}\n<\/style>\n<iframe src=\"\/labs\/electric-field.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\">\n<\/iframe>\n<\/div><\/div>\n\n<h2>Electric Field Lines and Uniform Fields<\/h2>\n\n<p>Field lines are the picture behind the maths. They are arrows that trace the direction a positive charge would move, and a handful of simple rules makes them reliable:<\/p>\n\n<ul>\n<li>Lines start on positive charges and end on negative charges (or run off to infinity).<\/li>\n<li>The field is stronger where the lines are <strong>closer together<\/strong>, weaker where they spread out.<\/li>\n<li>Lines never cross \u2014 the field can only point one way at any single point.<\/li>\n<li>Lines always meet a conductor&#8217;s surface at right angles.<\/li>\n<\/ul>\n\n<p>Around a lone point charge the lines fan out like spokes, getting sparser with distance \u2014 exactly the inverse-square fall-off, drawn. Around a pair of opposite charges they curve gracefully from positive to negative, forming the familiar dipole pattern.<\/p>\n\n<h3>The special case: a uniform field<\/h3>\n\n<p>Put two flat metal plates parallel to each other and connect them to a battery, and something neat happens. Between the plates the field becomes <strong>uniform<\/strong> \u2014 the same strength and direction everywhere, shown by evenly spaced parallel lines.<\/p>\n\n<p>For that uniform field, the formula is refreshingly simple:<\/p>\n\n<div class=\"pf-formula\">E = V \/ d<\/div>\n\n<ul>\n<li><strong>V<\/strong> \u2014 the potential difference (voltage) between the plates, in volts (V).<\/li>\n<li><strong>d<\/strong> \u2014 the separation between the plates, in metres (m).<\/li>\n<\/ul>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/electric-field-uniform-between-two-parallel-charged.webp\" width=\"360\" height=\"258\" alt=\"Uniform electric field between two parallel charged plates, shown as equally spaced parallel arrows pointing from the positive plate to the negative plate\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:180px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;\">Between parallel plates the field is uniform \u2014 equal everywhere \u2014 with strength E = V\/d.<\/p>\n\n<h2>Real-World Examples of Electric Fields<\/h2>\n\n<p>Electric fields are not a chalkboard abstraction. They are working components in everyday technology and in the natural world. Here are five you meet without noticing.<\/p>\n\n<h3>1. Laser printers and photocopiers<\/h3>\n\n<p>Inside every laser printer is a drum given a careful pattern of electric charge. The field around the charged regions grabs powdered toner exactly where the page needs ink, then presses it onto the paper. No field, no print.<\/p>\n\n<h3>2. Capacitors in your phone and camera flash<\/h3>\n\n<p>A capacitor is two plates with a uniform field between them, storing energy in that field. When a camera flash fires, a capacitor dumps its stored field energy in a millisecond. The same components smooth the power inside virtually every gadget you own.<\/p>\n\n<h3>3. Lightning<\/h3>\n\n<p>Storm clouds separate charge until the field between cloud and ground grows enormous. Once it passes roughly 3 million volts per metre, the air itself breaks down and conducts \u2014 and a lightning bolt is the field discharging in a fraction of a second.<\/p>\n<figure style=\"margin:32px auto;max-width:1200px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/5073_main_lightning-ground-discharge_electrometeors.jpeg\"\n       alt=\"Lightning discharging the electric field between cloud and ground\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"2313\" height=\"1542\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">A lightning bolt is the electric field between cloud and ground discharging in a fraction of a second.<\/figcaption>\n<\/figure>\n<h3>4. Touchscreens<\/h3>\n\n<p>A capacitive touchscreen holds a faint electric field across its surface. Your fingertip, being slightly conductive, distorts that field, and the phone reads exactly where the distortion happened. Every tap and swipe is a field measurement.<\/p>\n\n<h3>5. Van de Graaff generators (and hair-raising static)<\/h3>\n\n<p>Touch a Van de Graaff generator and your hair stands on end. Charge spreads over you and your strands, and because like charges repel along the field, each hair pushes away from its neighbours \u2014 a vivid, harmless demonstration of a field at work.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Situation<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Approximate field strength<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Earth&#8217;s fair-weather field near the ground<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">~100 N\/C<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Sunlight \/ radio waves reaching the Earth<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">~10<sup>2<\/sup>\u201310<sup>3<\/sup> N\/C<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Just before a spark jumps in dry air (breakdown)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">~3 \u00d7 10<sup>6<\/sup> V\/m<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Across a cell membrane in your body<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">~10<sup>7<\/sup> V\/m<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">At the electron&#8217;s orbit in a hydrogen atom<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">~5 \u00d7 10<sup>11<\/sup> N\/C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h2>Common Misconceptions About Electric Fields<\/h2>\n\n<h3>&#8220;The field only exists when there&#8217;s a test charge in it&#8221;<\/h3>\n\n<p>No. The source charge creates the field throughout the surrounding space all by itself. The test charge is just our <em>probe<\/em> for measuring what is already there \u2014 remove it and the field carries on existing.<\/p>\n\n<h3>&#8220;E = F\/q means the field depends on q&#8221;<\/h3>\n\n<p>It looks that way, but it isn&#8217;t. If you halve the test charge, the force on it also halves, so the ratio F\/q stays exactly the same. The field at a point is fixed by the source, not by whatever you use to measure it.<\/p>\n\n<h3>&#8220;Electric field and electric potential are the same thing&#8221;<\/h3>\n\n<p>They are partners, not twins. Field (E) is a vector measured in N\/C; potential (V) is a scalar measured in volts. The field is the rate at which potential changes with distance \u2014 steep potential means strong field. They are linked, but they are not interchangeable.<\/p>\n\n<h3>&#8220;A charge always travels along a field line&#8221;<\/h3>\n\n<p>Only sometimes. A charge released from rest does start moving along the field line through that point. But if it is already moving sideways, it follows a curved path that cuts across the lines \u2014 the field sets its acceleration, not its whole trajectory.<\/p>\n\n<h2>How Electric Fields Relate to Other Concepts<\/h2>\n\n<p>The electric field is a hub that connects much of electromagnetism. Seeing the links makes each topic easier.<\/p>\n\n<h3>Coulomb&#8217;s law and electric force<\/h3>\n\n<p>The field is Coulomb&#8217;s force, repackaged per unit charge. Multiply the field by a charge and you are straight back to a force: F = qE. If you want the force between two specific charges instead, <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/coulombs-law\/\">Coulomb&#8217;s law<\/a> is the tool.<\/p>\n\n<h3>Voltage, current and circuits<\/h3>\n\n<p>Inside a wire, it is an electric field that nudges the free electrons along, setting up a current. That field-driven flow is the starting point for <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/ohms-law\/\">Ohm&#8217;s law<\/a> and the whole of circuit theory.<\/p>\n\n<h3>Energy and work<\/h3>\n\n<p>Pushing a charge through a field takes work, and that work is stored as electric potential <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-energy-in-physics\/\">energy<\/a>. Letting the charge go converts it back into kinetic energy \u2014 the principle behind everything from a spark to a particle accelerator.<\/p>\n\n<h3>Magnetism and light<\/h3>\n\n<p>A changing electric field gives birth to a magnetic field, and a changing magnetic field gives birth to an electric one. Together they leapfrog through space as an electromagnetic wave \u2014 which is exactly why light travels at <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/modern-physics\/speed-of-light\/\">the speed of light<\/a>.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Quantity<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Symbol<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Type<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">SI unit<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Depends on\u2026<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Electric field<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">E<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Vector<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">N\/C (= V\/m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">source charge &amp; position (not the test charge)<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Electric force<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">F<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Vector<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">newton (N)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the field and the charge placed in it<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Electric potential<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">V<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Scalar<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">volt (V)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">source charge &amp; position<\/td>\n<\/tr>\n<tr style=\"background:#FAF6EE;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Potential energy<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">U<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Scalar<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">joule (J)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">the charge and the potential (U = qV)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A point charge of +2.0 nC sits in a vacuum. Find the electric field strength at a point 5.0 mm away.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: Use the point-charge formula, E = kQ\/r<sup>2<\/sup>, with k = 8.99 \u00d7 10<sup>9<\/sup> N\u00b7m<sup>2<\/sup>\/C<sup>2<\/sup>.<\/p>\n\n<p>Step 2: Convert units: Q = 2.0 \u00d7 10<sup>\u22129<\/sup> C, r = 5.0 \u00d7 10<sup>\u22123<\/sup> m, so r<sup>2<\/sup> = 2.5 \u00d7 10<sup>\u22125<\/sup> m<sup>2<\/sup>.<\/p>\n\n<p>Step 3: E = (8.99 \u00d7 10<sup>9<\/sup> \u00d7 2.0 \u00d7 10<sup>\u22129<\/sup>) \u00f7 (2.5 \u00d7 10<sup>\u22125<\/sup>) = 17.98 \u00f7 (2.5 \u00d7 10<sup>\u22125<\/sup>).<\/p>\n\n<p><strong>Answer: E \u2248 7.2 \u00d7 10<sup>5<\/sup> N\/C, directed radially outward (the charge is positive).<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A charge of 5.0 \u03bcC is placed at a point where the electric field is 2.0 \u00d7 10^3 N\/C. What force does it feel?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: The force on a charge in a field is F = qE.<\/p>\n\n<p>Step 2: Substitute q = 5.0 \u00d7 10<sup>\u22126<\/sup> C and E = 2.0 \u00d7 10<sup>3<\/sup> N\/C.<\/p>\n\n<p>Step 3: F = (5.0 \u00d7 10<sup>\u22126<\/sup>)(2.0 \u00d7 10<sup>3<\/sup>) = 1.0 \u00d7 10<sup>\u22122<\/sup> N.<\/p>\n\n<p><strong>Answer: F = 0.010 N, in the direction of the field.<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">At what distance from a +1.0 nC charge is the electric field equal to 900 N\/C?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: Start from E = kQ\/r<sup>2<\/sup> and rearrange for r: r = \u221a(kQ\/E).<\/p>\n\n<p>Step 2: Substitute: r = \u221a[(8.99 \u00d7 10<sup>9<\/sup> \u00d7 1.0 \u00d7 10<sup>\u22129<\/sup>) \u00f7 900] = \u221a(8.99 \u00f7 900).<\/p>\n\n<p>Step 3: r = \u221a(9.99 \u00d7 10<sup>\u22123<\/sup>) = 0.0999 m.<\/p>\n\n<p><strong>Answer: r \u2248 0.10 m (about 10 cm).<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">An electron (charge 1.6 \u00d7 10^-19 C, mass 9.11 \u00d7 10^-31 kg) is placed in a uniform field of 1.0 \u00d7 10^4 N\/C. Find its acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: Find the force with F = qE.<\/p>\n\n<p>Step 2: F = (1.6 \u00d7 10<sup>\u221219<\/sup>)(1.0 \u00d7 10<sup>4<\/sup>) = 1.6 \u00d7 10<sup>\u221215<\/sup> N.<\/p>\n\n<p>Step 3: Use Newton&#8217;s second law, a = F\/m = (1.6 \u00d7 10<sup>\u221215<\/sup>) \u00f7 (9.11 \u00d7 10<sup>\u221231<\/sup>).<\/p>\n\n<p><strong>Answer: a \u2248 1.8 \u00d7 10<sup>15<\/sup> m\/s<sup>2<\/sup>, directed opposite to the field (the electron is negative).<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">Two parallel plates are 4.0 mm apart with a potential difference of 120 V across them. Find the field between them, and the force on a 2.0 nC charge placed there.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: For a uniform field, E = V\/d, with d = 4.0 \u00d7 10<sup>\u22123<\/sup> m.<\/p>\n\n<p>Step 2: E = 120 \u00f7 (4.0 \u00d7 10<sup>\u22123<\/sup>) = 3.0 \u00d7 10<sup>4<\/sup> V\/m (= 3.0 \u00d7 10<sup>4<\/sup> N\/C).<\/p>\n\n<p>Step 3: Force: F = qE = (2.0 \u00d7 10<sup>\u22129<\/sup>)(3.0 \u00d7 10<sup>4<\/sup>) = 6.0 \u00d7 10<sup>\u22125<\/sup> N.<\/p>\n\n<p><strong>Answer: E = 3.0 \u00d7 10<sup>4<\/sup> N\/C; F = 6.0 \u00d7 10<sup>\u22125<\/sup> N.<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A +8.0 nC charge sits at x = 0 and a \u22128.0 nC charge at x = 0.40 m. Find the net electric field at the midpoint, x = 0.20 m.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: Each charge is 0.20 m from the midpoint. Find each field magnitude with E = kQ\/r<sup>2<\/sup>.<\/p>\n\n<p>Step 2: E = (8.99 \u00d7 10<sup>9<\/sup> \u00d7 8.0 \u00d7 10<sup>\u22129<\/sup>) \u00f7 (0.20)<sup>2<\/sup> = 71.92 \u00f7 0.04 = 1798 N\/C from each charge.<\/p>\n\n<p>Step 3: At the midpoint, the field from the +8.0 nC charge points toward +x, and the field toward the \u22128.0 nC charge <em>also<\/em> points toward +x, so they add: 1798 + 1798.<\/p>\n\n<p><strong>Answer: E \u2248 3.6 \u00d7 10<sup>3<\/sup> N\/C, pointing from the positive charge toward the negative charge.<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A tiny oil drop of mass 4.0 \u00d7 10^-15 kg hangs motionless in a vertical field of 8.2 \u00d7 10^4 N\/C. What charge does it carry? (Take g = 9.81 m\/s^2.)<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: &#8220;Motionless&#8221; means the electric force balances gravity: qE = mg.<\/p>\n\n<p>Step 2: Rearrange: q = mg\/E = (4.0 \u00d7 10<sup>\u221215<\/sup> \u00d7 9.81) \u00f7 (8.2 \u00d7 10<sup>4<\/sup>).<\/p>\n\n<p>Step 3: q = (3.92 \u00d7 10<sup>\u221214<\/sup>) \u00f7 (8.2 \u00d7 10<sup>4<\/sup>) = 4.8 \u00d7 10<sup>\u221219<\/sup> C.<\/p>\n\n<p><strong>Answer: q \u2248 4.8 \u00d7 10<sup>\u221219<\/sup> C \u2014 almost exactly 3 electron charges, echoing Millikan&#8217;s famous experiment.<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A +9.0 nC charge is at x = 0 and a +4.0 nC charge at x = 1.0 m. Where on the line between them is the net field zero?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<p><strong>Solution:<\/strong><\/p>\n\n<p>Step 1: Between two positive charges the fields point in opposite directions, so they can cancel. Set the magnitudes equal at distance x from the +9.0 nC charge: kQ<sub>1<\/sub>\/x<sup>2<\/sup> = kQ<sub>2<\/sub>\/(1 \u2212 x)<sup>2<\/sup>.<\/p>\n\n<p>Step 2: The k cancels, leaving Q<sub>1<\/sub>\/x<sup>2<\/sup> = Q<sub>2<\/sub>\/(1 \u2212 x)<sup>2<\/sup>, so (1 \u2212 x)\/x = \u221a(Q<sub>2<\/sub>\/Q<sub>1<\/sub>) = \u221a(4\/9) = 2\/3.<\/p>\n\n<p>Step 3: Then 1 \u2212 x = (2\/3)x, so 1 = (5\/3)x, giving x = 0.60 m.<\/p>\n\n<p><strong>Answer: The field is zero 0.60 m from the +9.0 nC charge (0.40 m from the +4.0 nC charge).<\/strong><\/p>\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is an electric field in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nAn electric field is the invisible &#8220;zone of influence&#8221; around a charged object that pushes or pulls on other charges. At every point it tells you the force a positive charge would feel there. The field exists in the space itself, whether or not another charge is present to experience it.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the formula for electric field?<\/summary><div class=\"pf-faq-item-answer\">\n\nThere are two key formulas. The definition is E = F\/q \u2014 the force on a test charge divided by that charge. For the field created by a single point charge, use E = kQ\/r<sup>2<\/sup>, where k \u2248 8.99 \u00d7 10<sup>9<\/sup> N\u00b7m<sup>2<\/sup>\/C<sup>2<\/sup>, Q is the source charge and r is the distance. Both give the field in newtons per coulomb.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What are the units of electric field?<\/summary><div class=\"pf-faq-item-answer\">\n\nElectric field is measured in newtons per coulomb (N\/C). An identical unit is volts per metre (V\/m), because a volt is a joule per coulomb and a joule is a newton-metre. Use N\/C when thinking about force on a charge, and V\/m when thinking about voltage across a gap.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is electric field a vector or a scalar?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe electric field is a vector \u2014 it has both a magnitude and a direction at every point. The direction is defined as the way a positive test charge would be pushed: away from positive charges and toward negative ones. When several charges are present, you add their field vectors, not just their sizes.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between electric field and electric force?<\/summary><div class=\"pf-faq-item-answer\">\n\nElectric force (F) acts on a specific charge and is measured in newtons; remove the charge and the force vanishes. Electric field (E) is force per unit charge and exists in the space whether or not a charge is there. They are linked by F = qE \u2014 multiply the field by a charge to get the force.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the electric field depend on the test charge?<\/summary><div class=\"pf-faq-item-answer\">\n\nNo. Although E = F\/q contains the test charge q, the force F is proportional to q, so the ratio stays constant. The field at any point is set entirely by the source charges and the distance from them. A larger probe feels a larger force, but it measures the same field.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>A clear, complete guide to the electric field: what it is, the two key formulas (E = F\/q and E = kQ\/r\u00b2), how to read field lines, five real-world examples and worked problems.<\/p>\n","protected":false},"author":1,"featured_media":383,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[68,222,223,70,69],"class_list":["post-382","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electromagnetism","tag-coulombs-law","tag-electric-field","tag-electric-field-formula","tag-electromagnetism","tag-electrostatics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/382","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=382"}],"version-history":[{"count":6,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/382\/revisions"}],"predecessor-version":[{"id":1663,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/382\/revisions\/1663"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/383"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=382"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=382"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=382"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}